Question

What mass of CO2 is produced when 25.00 mL sample of ethanol (density = 0.789 g/mL) combusts?

C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(l)

Answer 1

m CO2 = 37.686 g

Step-by-step explanation:

• C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)

∴ V ethanol = 25.00 mL

∴ δ ethanol = 0.789 g/mL

∴ Mw ethanol = 46.07 g/mol

∴ Mw CO2 = 44.01 g/mol

⇒ m ethanol = (0.789 g/mL)×(25.00mL) = 19.725 g C2H5OH(l)

⇒ m CO2(g) = (19.725 g C2H5OH)×(mol C2H5OH/46.07 gC2H5OH)×(2mol CO2/mol C2H5OH)×(44.01 g CO2/mol CO2)

⇒ m CO2 = 37.686 g CO2